Discuss the shape of the following molecules using the VSEPR model: BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃.

Exercises · 4.7

BeCl_2

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Concept understanding — Lewis Dot Structures

Why do we need Lewis dot structures?

Atoms are held together in molecules by chemical bonds — but what exactly is a bond? In the early 20th century, Gilbert N. Lewis realised that the key lies in the valence electrons (the outermost electrons). He noticed that atoms of noble gases (like Ne, Ar) are extremely stable and unreactive, and they all have 8 electrons in their outermost shell (except helium, which has 2). This led to the octet rule: atoms tend to gain, lose, or share electrons to achieve a full outer shell of 8 electrons (or 2 for hydrogen).

Lewis dot structures are simply a shorthand picture of this idea. They show:

Which atoms are connected to which
How many valence electrons each atom contributes
How those electrons are arranged as bonding pairs (shared) or lone pairs (unshared)

The precise statement

A Lewis dot structure (or electron dot structure) represents the valence electrons of an atom or molecule using dots placed around the element's symbol. Each dot stands for one valence electron. Shared pairs (bonds) are shown as lines, and unshared pairs as pairs of dots.

For a single atom, you write the element symbol and place dots on its four sides (top, bottom, left, right) — up to 8 dots. The order of filling doesn't matter for the final picture, but conventionally you place one dot on each side first, then pair them up.

For example:

Carbon (group 14, 4 valence electrons): $\cdot \text{C} \cdot$ (four single dots)
Oxygen (group 16, 6 valence electrons): $\cdot \ddot{\text{O}} \cdot$ (two single dots and two pairs)

How to draw a Lewis structure for a molecule

Here's the step-by-step method you'll use in exams:

Count total valence electrons — add up valence electrons from all atoms. For ions, add 1 electron for each negative charge, subtract 1 for each positive charge.
Identify the central atom — usually the least electronegative element (not hydrogen or fluorine). Place it in the centre.
Connect atoms with single bonds — each bond uses 2 electrons. Subtract these from your total.
Complete octets of outer atoms — place remaining electrons as lone pairs on terminal atoms (except hydrogen, which only needs 2).
Place leftover electrons on the central atom — if any remain.
If the central atom has fewer than 8 electrons, form multiple bonds — move lone pairs from outer atoms to create double or triple bonds until the central atom has an octet.

Watch out

A common mistake: forgetting that hydrogen only needs 2 electrons (a duet), not 8. Never put more than 2 electrons around H.

A concrete example: water (H₂O)

Total valence electrons: O has 6, each H has 1 → $6 + 1 + 1 = 8$ electrons.
Central atom: oxygen (least electronegative after H).
Connect: O—H bonds (2 bonds × 2 electrons = 4 electrons used).
Remaining: $8 - 4 = 4$ electrons → place as two lone pairs on oxygen.
Check: O has 2 bonds (4 electrons) + 2 lone pairs (4 electrons) = 8. Each H has 1 bond (2 electrons) = 2. Done.

The structure: $\text{H} - \ddot{\text{O}} - \text{H}$

What the structure tells you

Once drawn, a Lewis structure reveals:

Bond order (single, double, triple)
Lone pairs (which affect molecular shape and reactivity)
Formal charge (a bookkeeping tool to check which structure is most stable) …

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Concept: VSEPR Theory – electron groups (bonds + lone pairs) repel to minimise energy, determining molecular shape.

Reasoning steps:

Count valence electrons and draw the Lewis structure for each molecule.
Determine the steric number (SN = number of bonded atoms + lone pairs on the central atom).
Use SN to predict electron‑pair geometry, then adjust for lone pairs to get the molecular shape.

Molecule	Central atom	SN	Electron‑pair geometry	Molecular shape
$BeCl_2$	Be	2	Linear	Linear
$BCl_3$	B	3	Trigonal planar	Trigonal planar
$SiCl_4$	Si	4	Tetrahedral	Tetrahedral

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Using the VSEPR model, the shape of a molecule is determined by the number of bonding and lone pairs around the central atom. $BeCl_2$ is linear, $BCl_3$ is trigonal planar, $SiCl_4$ is tetrahedral, $AsF_5$ is trigonal bipyramidal, $H_2S$ is bent, and $PH_3$ is trigonal pyramidal.

The VSEPR (Valence Shell Electron Pair Repulsion) model is built on a simple, powerful idea: electron pairs — whether they are in bonds or lone pairs — repel each other. They arrange themselves as far apart as possible around a central atom to minimize this repulsion. The shape of the molecule is then determined by the positions of the atoms (not the lone pairs), but the lone pairs still push the bonds out of the way.

Let’s walk through each molecule step by step.

1. $BeCl_2$

Step 1: Count valence electrons.

Beryllium has 2 valence electrons. Each chlorine has 7, but we only care about the central atom’s electron count for the basic VSEPR geometry. Be forms two single bonds with the two Cl atoms.

Step 2: Determine electron pairs around Be.

Be has 2 bonding pairs and no lone pairs. That’s a total of 2 electron domains.

Step 3: Predict geometry.

Two electron domains repel to opposite sides — the angle is $180^\circ$ . The shape is linear.

Tip

Be is an exception to the octet rule — it’s happy with just 4 electrons (two bonds). This is why $BeCl_2$ is linear and stable.

2. $BCl_3$

Step 1: Central atom electron count.

Boron has 3 valence electrons. It forms three single bonds with three Cl atoms.

Step 2: Electron domains.

3 bonding pairs, 0 lone pairs. Total = 3 domains.

Step 3: Geometry.

Three domains repel to the corners of an equilateral triangle — bond angles $120^\circ$ . The shape is trigonal planar.

Watch out

A common mistake is to think boron “needs” an octet. It doesn’t — it’s electron-deficient and stable with 6 electrons. So no lone pairs, no bending.

3. $SiCl_4$

Step 1: Central atom.

Silicon has 4 valence electrons. It forms four single bonds with four Cl atoms.

Step 2: Electron domains.

4 bonding pairs, 0 lone pairs. Total = 4 domains.

Step 3: Geometry.

Four domains repel to the vertices of a tetrahedron — bond angles $109.5^\circ$ . The shape is tetrahedral.

Note

This is the classic “perfect” tetrahedron. All bonds are equivalent, and the molecule is nonpolar despite the polar bonds, because symmetry cancels the dipoles.

4. $AsF_5$

Step 1: Central atom.

Arsenic has 5 valence electrons. It forms five single bonds with five F atoms.

Step 2: Electron domains.

5 bonding pairs, 0 lone pairs. Total = 5 domains.

Step 3: Geometry.

Five domains arrange as a trigonal bipyramid. There are two distinct positions: three equatorial bonds (at $120^\circ$ to each other) and two axial bonds (at $90^\circ$ to the equatorial plane). The shape is trigonal bipyramidal.

Tip

In a trigonal bipyramid, the axial bonds are slightly longer than the equatorial ones because they experience more repulsion from the equatorial pairs. This is a key detail for advanced questions.

5. $H_2S$

Step 1: Central atom.

Sulfur has 6 valence electrons. It forms two single bonds with two H atoms.

Step 2: Electron domains.

2 bonding pairs, 2 lone pairs. Total = 4 domains.

Step 3: Geometry.

Four domains would ideally be tetrahedral ( $109.5^\circ$ ), but lone pairs repel more strongly than bonding pairs. They push the H–S–H bond angle down to about $92^\circ$ (experimentally). The shape is bent (or V-shaped). …

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Q.Discuss the shape of the following molecules using the VSEPR model: BeCl2BeCl_2BeCl2​, BCl3BCl_3BCl3​, SiCl4SiCl_4SiCl4​, AsF5AsF_5AsF5​, H2SH_2SH2​S, PH3PH_3PH3​.

1. BeCl2BeCl_2BeCl2​

2. BCl3BCl_3BCl3​

3. SiCl4SiCl_4SiCl4​

4. AsF5AsF_5AsF5​

5. H2SH_2SH2​S

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Q.Discuss the shape of the following molecules using the VSEPR model: $BeCl_2$ , $BCl_3$ , $SiCl_4$ , $AsF_5$ , $H_2S$ , $PH_3$ .

1. $BeCl_2$

2. $BCl_3$

3. $SiCl_4$

4. $AsF_5$

5. $H_2S$

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